Sixty four conducting drops each of radius 0 . 02   m and each carrying a charge of 5   μ C are combined to form a bigger drop. The ratio of surface density of bigger drop to the smaller drop will be

Question

Sixty four conducting drops each of radius 0 . 02 &#160; m and each carrying a charge of 5 &#160; &#956; C are combined to form a bigger drop. The ratio of surface density of bigger drop to the smaller drop will be

Aicharya · Accepted Answer

Total volume will be constant. Therefore, $n (\frac{4 π}{3} r^{3}) = \frac{4 π}{3} R^{3} \Rightarrow 64^{\frac{1}{3}} r = R \Rightarrow R = 4 r$

Final surface charge density $σ^{'} = \frac{n σ_{0} 4 π r^{2}}{4 π R^{2}} = \frac{64 \times σ_{0} r^{2}}{16 r^{2}} \Rightarrow \frac{σ^{'}}{σ_{0}} = \frac{4}{1}$

Sixty four conducting drops each of radius 0 . 02   m and each carrying a charge of 5   μ C…

Solution

Sixty four conducting drops each of radius 0 . 02 &#160; m and each carrying a charge of 5 &#160; &#956; C…

Solution

Sixty four conducting drops each of radius 0 . 02 m and each carrying a charge of 5 μ C…